The maximum electric power required in a city with a random variable X is modeled has a probability density function Below:

$$f(x) = c^{2}xe^{-cx}$$ $$c= 5\times10^{-6} per kilowatt$$ First if the available power is 1 million kilowatts, what is the probability of blackout?

Second what is the amount of power needed so that the probability of blackout to be less than 0.005?

how can we calculate the blackout probability?

  • 1
    $\begingroup$ Without an explicit expression for $U(x)$ the questions can't be answered. $\endgroup$ – herb steinberg Oct 14 '18 at 21:29
  • $\begingroup$ Exactly what i was thinking. $\endgroup$ – lighting Oct 15 '18 at 9:31
  • $\begingroup$ What if we ignore $U(x)$, how we can solve it? $\endgroup$ – lighting Oct 16 '18 at 17:39

Step: the cumulative distribution $F(x)=1-e^{-cx}(1+cx)$

For the first question, $cx=5$, blackout probability is $1-F(x)=(1+cx)e^{-cx}=.0337$.

For the second question, let $1-F(x)=.005$, or $(1+cx)e^{-cx}=.005$. Solve for $cx$. Numerical method needed for solution. I got $cx=7.425$ or $x=1.485$ million kilowatts.

  • $\begingroup$ How is $cx =5$ ? the available power in question was $x$? $\endgroup$ – lighting Oct 16 '18 at 19:25
  • $\begingroup$ $c=5\times 10^{-6},\ x=10^6$. Those are the numbers you gave. $x$= available power. $\endgroup$ – herb steinberg Oct 16 '18 at 21:26

There is an analytical solution for the equation $$(1+cx)e^{-cx}=k$$ Rewrite it as $$(1+cx)e^{-(cx+1)}=\frac ke$$ Let $cx+1=t$ to get $$t\,e^{-t}=\frac ke\implies t=-W_{-1}\left(-\frac{k}{e}\right)$$ where appears Lambert function.

Back to $x$, the solution is then $$x=-\frac{W_{-1}\left(-\frac{k}{e}\right)+1}{c}$$ Use the formula given in the linked Wikipedia page $$W(t)=L_1-L_2+\frac{L_2}{L_1}+\frac{L_2(L_2-2)}{2L_1^2}+\frac{L_2(6-9L_2+2L_2^2)}{6L_1^3}+\cdots$$ where $L_1=\log(-t)$ and $L_2=\log(-L_1)$ and you will get $x=1.48603\times 10^6$.


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